probability of a water molecule in its flexing ground state

A water molecule can vibrate in various ways, but the easiest type of vibration to excite is the "flexing" mode in which the hydrogen atoms move toward and away from each other by the HO bonds do not stretch.

The oscillations of this mode are approximately harmonic, with a frequency of 4.8 x 10^13 Hz. As for any quantum harmonic oscillator, the energy levels are ½ hf, 3/2 hf, 5/2 hf, and so on. None of these levels are degenerate.

(a) Calculate the probability of a water molecule being in its flexing ground state and in each of the first two excited states, assuming that it is in equilibrium with a reservoir (say the atmosphere) at 300 K. (Hint: Calculate Z by adding up the first few Boltzmann factors, until the rest are negligible.)

(b) Repeat the calculation for a water molecule in equilibrium with a reservoir at 700 K (perhaps in a steam turbine).

Solution Summary

It calculates the probability of a water molecule being in its flexing ground state and in each of the first two excited states at two different temperatures 300 K and 700 K, respectively. The solution is detailed and was rated '5/5' by the student who posted the question.